Revision as of 13:33, 13 December 2010

Practice Question 4, ECE438 Fall 2010, Prof. Boutin

Frequency domain view of filtering.

Note: There is a very high chance of a question like this on the final.

Define a signal x(t) and take samples every T (using a specific value of T). Store the samples in a discrete-time signal z[n]. Obtain a mathematical expression for the Fourier transform of x(t) and sketch it. Obtain a mathematical expression for the Fourier transform of y[n] and sketch it.

Let's hope we get a lot of different signals from different students!

• Instructor's note: if you make a mistake and somebody else comments on it, please do not remove it when you make corrections later. Instead, just start over below your previous answer. The idea is to learn from our mistakes! -pm

Post Your answer/questions below.

I thought I would start with a function that had a simple F.T.

$ x(t) = \delta(t), T=1 $

$ \begin{align} z[n] &= x_T[n] \\ &= \delta(t+T) \end{align} $

Fourier Transform of x(t) = 1

$ y[n] = x(t)*z[n] $ <-- is this correct?

• No, there is a serious problem with your signal y[n]: it's defined as a function of both t and n. It should NOT depend on n.

I only solved the general form for this problem.

$ \begin{align} \mathcal{F}(x(t)) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} \end{align} $

$ \begin{align} z[n] = comb_T(x(t)) \end{align} $

• Oops! Your z[n] is actually a function of t (because a comb is the multiplication of two functions of t, namely the signal x(t) and an impulse train p(t) .

I'm not sure what y[n] is equal to. I'm assuming that y[n] is the same as z[n]. Then the FT of y[n] is

$ \begin{align} Y(e^{j\omega}) = \frac{1}{T}rep_\frac{1}{T}(X(e^{j\omega})) \end{align} $

- Mike Wolfer

T=3

$ \begin{align} x(t)=cos(2t) \end{align} $

Sampling consists mathematically of a substitution of t->3n Therefore,

$ \begin{align} z(n) = cos(2(3n)) \end{align} $

To make calculations easier Euler Form is used:

$ \begin{align} cos(6n) -> 1/2(e^{i6n}+e^{-i6n}) \end{align} $

Inverse CTFT:

$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \! X(\omega)e^{j \omega t} dw $

$ 1/2e^{i2t}+1/2e^{-i2t} = \frac{1}{2\pi} \int_{-\infty}^{\infty} \! X(\omega)e^{j \omega t} dw $

Table -> $ \begin{align} x(f)=(\delta(f - 2/(2\pi)) +\delta(f + 2/(2\pi))) \end{align} $

Sketch consists of two positive spikes, at 1/pi and -1/pi on the X axis(f).

IDFT:

$ x[n] = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j \omega n} d\omega $

$ 1/2e^{i6n}+1/2e^{-i6n}= \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j \omega n} d\omega $

Table ->

$ \begin{align} z(w)=\pi( \delta(\omega -6) +\delta(\omega+6) ) \end{align} $ Sketch consists of a pair of spikes repeated every 2 pi. Given that the original is not in the initial +/-pi range, it is shifted by 2 pi intervals till the spikes are within.

6-2pi-> spikes at +/- .283

• Answer/question

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